(solved)Question 2.23 of NCERT Class XI Chemistry Chapter 2

$\displaystyle \underline{\underline{\text{Part (i)}}}$

calculate electrons first, then configure

$\displaystyle \text{(a) }Z_H = 1\therefore n_e \text{ in H}^-$ = 1 + 1 = 2

$\displaystyle \implies 1s^2 \:\underline{Ans}$

$\displaystyle \text{(b) }Z_{Na} = 11\therefore n_e \text{ in Na}^+$ = 11 - 1 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \text{(c) }Z_{O} = 8\therefore n_e \text{ in O}^{2-}$ = 8 + 2 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \text{(d) }Z_{F} = 9\therefore n_e \text{ in F}^{-}$ = 9 + 1 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (ii)}}}$

Complete the previous shells in each case

$\displaystyle \text{(a) }3s^1 \equiv 1s^22s^22p^63s^1 \implies Z = 11$

$\displaystyle \text{(b) }2p^3 \equiv 1s^22s^22p^3 \implies Z = 7$

$\displaystyle \text{(c) }3p^5 \equiv 1s^22s^22p^63s^23p^5 \implies Z = 17$

$\displaystyle \underline{\underline{\text{Part (iii)}}}$

Calculate Z in each case

$\displaystyle \text{(a) [He]}2s^1 $

$\displaystyle \therefore Z = Z_{He} + 1 = 2 + 1 = 3 \text{ Li }\underline{Ans}$

$\displaystyle \text{(b) [Ne]}3s^23p^3 $

$\displaystyle \therefore Z = Z_{Ne} + 5 = 10 + 5 = 15 \text{ P }\underline{Ans}$

$\displaystyle \text{(c) [Ar]}4s^23d^1 $

$\displaystyle \therefore Z = Z_{Ar} + 3 = 18 + 3 = 21 \text{ Sc }\underline{Ans}$